\subsection{Conjugation in alternating groups}
We know that elements in \( S_n \) are conjugate if and only if they have the same cycle type.
However, elements of \( A_n \) that are conjugate in \( S_n \) are not necessarily conjugate in \( A_n \).
Let \( g \in A_n \).
Then \( C_{A_n}(g) = C_{S_n}(g) \cap A_n \).
There are two possible cases.
\begin{itemize}
	\item If there exists an odd permutation that commutes with \( g \), then \( 2\abs{C_{A_n}}(g) = \abs{C_{S_n}}(g) \).
	      By the orbit-stabiliser theorem, \( \abs{\ccl_{A_n}(g)} = \abs{\ccl_{S_n}(g)} \).
	\item If there is no odd permutation that commutes with \( g \), we have \( \abs{C_{A_n}}(g) = \abs{C_{S_n}}(g) \).
	      Similarly, \( 2\abs{\ccl_{A_n}(g)} = \abs{\ccl_{S_n}(g)} \).
\end{itemize}
\begin{example}
	For \( n = 5 \), the product \( (1\ 2)(3\ 4) \) commutes with \( (1\ 2) \), and \( (1\ 2\ 3) \) commutes with \( (4\ 5) \).
	Both of these elements are odd.
	So the conjugacy classes of the above inside \( S_5 \) and \( A_5 \) are the same.
	However, \( (1\ 2\ 3\ 4\ 5) \) does not commute with any odd permutation.
	Indeed, if that were true for some \( h \), we would have
	\[
		(1\ 2\ 3\ 4\ 5) = h (1\ 2\ 3\ 4\ 5) h^{-1} = (h(1)\ h(2)\ h(3)\ h(4)\ h(5))
	\]
	Hence \( h \) must be a 5-cycle in the subgroup of \( A_5 \) generated by \( (1\ 2\ 3\ 4\ 5) \).

	We can then show that \( A_5 \) has conjugacy classes of size \( 1, 15, 20, 12, 12 \).
	If \( H \trianglelefteq A_5 \), \( \abs{H} \) must be a sum of the sizes of the above conjugacy classes.
	By Lagrange's theorem, \( \abs{H} \) must divide 60.
	We can check explicitly that this is not possible unless \( \abs{H} = 1 \) or \( \abs{H} = 60 \).
	Hence \( A_5 \) is simple.
\end{example}

\subsection{Simplicity of alternating groups}
\begin{lemma}
	\( A_n \) is generated by 3-cycles.
\end{lemma}
\begin{proof}
	All elements of \( A_n \) are generated by an even number of transpositions.
	It therefore suffices to show that a product of two transpositions can be written as a product of 3-cycles.
	Explicitly,
	\[
		(a\ b)(c\ d) = (a\ c\ b)(a\ c\ d);\quad (a\ b)(b\ c) = (a\ b\ c)
	\]
\end{proof}
\begin{lemma}
	If \( n \geq 5 \), all 3-cycles in \( A_n \) are conjugate (in \( A_n \)).
\end{lemma}
\begin{proof}
	We claim that every 3-cycle is conjugate to \( (1\ 2\ 3) \).
	If \( (a\ b\ c) \) is a 3-cycle, we have \( (a\ b\ c) = \sigma(1\ 2\ 3)\sigma^{-1} \) for some \( \sigma \in S_n \).
	If \( \sigma \in A_n \), then the proof is finished.
	Otherwise, \( \sigma \mapsto \sigma(4\ 5) \in A_n \) suffices, since \( (4\ 5) \) commutes with \( (1\ 2\ 3) \).
\end{proof}
\begin{theorem}
	\( A_n \) is simple for \( n \geq 5 \).
\end{theorem}
\begin{proof}
	Suppose \( 1 \neq N \vartriangleleft A_n \).
	To disprove normality, it suffices to show that \( N \) contains a 3-cycle by the lemmas above, since the normality of \( N \) would imply \( N \) contains all 3-cycles and hence all elements of \( A_n \).

	Let \( 1 \neq \sigma \in N \), writing \( \sigma \) as a product of disjoint cycles.
	\begin{enumerate}
		\item Suppose \( \sigma \) contains a cycle of length \( r \geq 4 \).
		      Without loss of generality, let \( \sigma = (1\ 2\ 3\dots r) \tau \) where \( \tau \) fixes \( 1, \dots, r \).
		      Now, let \( \delta = (1\ 2\ 3) \).
		      We have
		      \[
			      \underbrace{\sigma^{-1}}_{\in N} \underbrace{\delta^{-1} \sigma \delta}_{\in N} = (r \dots 2\ 1)(1\ 3\ 2)(1\ 2\dots r) = (2\ 3\ r)
		      \]
		      So \( N \) contains a 3-cycle.
		\item Suppose \( \sigma \) contains two 3-cycles, which can be written without loss of generality as \( (1\ 2\ 3)(4\ 5\ 6) \tau \).
		      Let \( \delta = (1\ 2\ 4) \), and then
		      \[
			      \sigma^{-1} \delta^{-1} \sigma \delta = (1\ 3\ 2)(4\ 6\ 5)(1\ 4\ 2)(1\ 2\ 3)(4\ 5\ 6)(1\ 2\ 4) = (1\ 2\ 4\ 3\ 6)
		      \]
		      Therefore, there exists an element of \( N \) which contains a cycle of length \( 5 \geq 4 \).
		      This reduces the problem to case (i).
		\item Finally, suppose \( \sigma \) contains two 2-cycles, which will be written \( (1\ 2)(3\ 4)\tau \).
		      Then let \( \delta = (1\ 2\ 3) \) and
		      \[
			      \sigma^{-1} \delta^{-1} \sigma \delta = (1\ 2)(3\ 4)(1\ 3\ 2)(1\ 2)(3\ 4)(1\ 2\ 3) = (1\ 4)(2\ 3) = \pi
		      \]
		      Let \( \varepsilon = (2\ 3\ 5) \).
		      Then
		      \[
			      \underbrace{\pi^{-1}}_{\in N} \underbrace{\varepsilon^{-1} \pi \varepsilon}_{\in N} = (1\ 4)(2\ 3)(2\ 5\ 3)(1\ 4)(2\ 3)(2\ 3\ 5) = (2\ 5\ 3)
		      \]
		      Thus \( N \) contains a 3-cycle.
	\end{enumerate}
	There are now three remaining cases, where \( \sigma \) is a transposition, a 3-cycle, or a transposition composed with a 3-cycle.
	Note that the remaining cases containing transpositions cannot be elements of \( A_n \).
	If \( \sigma \) is a 3-cycle, we already know \( A_n \) contains a 3-cycle, namely \( \sigma \) itself.
\end{proof}
